Does the property $\forall x\exists y\forall z: x*y*z=e$ hold in a group?

Question

Does the given property hold in a group. If yes, why?

$$\forall x\exists y\forall z : x*y*z=e; \quad x,y,z\in G$$

It points out:

1. all elements (given by $z$ above) have unique inverse (given by $xy = z^{-1}$ above).
2. Set of inverse of elements of group = set of group elements.
3. Each row has the complete set.

Can use left cancellation to show row elements are unique, as given two entries in group table: $ab = ac \implies b=c$.

Similarly, for uniqueness of column elements can use right-cancellation. There $ba= ca\implies a=c$.

But, that is not the correct approach, as need to first find reason for #1.

Answer 1

The stated property $$\forall x \exists y\forall z : x*y*z = e$$ is true if and only if $G = \{e\}$ is the trivial group.

Set $x = e$. Then the question is if there is a $y$ with $y*z = e$ for all $z$. Now $z = y^{-1}$ is uniquely determined by $y$, so the property is wrong as soon as $G$ contains $2$ different elements.

Addition In the comments, the alternative property $$\forall x\forall z\exists y: x*y*z = e$$ has been discussed. The crucial difference to the original one is that now $y$ may be specified after $z$ has been fixed, so we can pick $y$ depending on $x$ and $z$ (and not only depending on $x$ as before). Consequently, the alternative property is softer than the original one. Indeed, it is true for all groups $G$ (setting $y = x^{-1} z^{-1}$).

Answer 2

(Adapted from a comment)

Be very careful with the order of quantifiers. Here's a tip: any time you have lots of quantifiers, you can think of it as a game.

Take the statement $\forall x\exists y\forall z:xyz=e$. This means: I choose an $x$, you choose a $y$, and I choose a $z$. I'm trying to make $xyz=e$ false. You're trying to make $xyz=e$ true. The statement $\forall x\exists y\forall z:xyz=e$ is true if and only if you have a winning strategy in this game. ("For any move I make, there is a move you can make so that for any move I make, you win.")

Here, assuming $G$ is not the trivial group (it has at least two elements), you do not have a winning strategy in this game: I have the final move and so I can muck it up at the last second. Example: I pick $x=e$, you pick some $y$, I pick $z=y^{-1}g$ where $g\ne e$.

If you change it to $\forall x\forall z\exists y$, the game changes to me choosing $x$ and $z$ and then you choosing $y$. Now, you do have a winning strategy because after I choose $x$ and $z$, you can choose $y=x^{-1}z^{-1}$ and win.

$\forall$ means I move. $\exists$ means you move. Changing the order of quantifiers means changing the order of who moves in the game.